GRE Mathematics Subject
Test GR0568 Solutions
2nd edition
Charles Rambo
Preface
Thank you for checking out my solutions to the GRE mathematics subject test 0568. This is primarily
intended as a reference, though some theorems were stated only once and then simply referenced thereafter
for brevity. I tried to reference major theorems when they were used, especially those from Calculus since I
imagine it has been awhile for a lot of you. I would like to acknowledge how much Wikipedia has helped me
refine the theorems I’ve stated throughout this document. If you need the original GRE subject test 0568
questions, go to http://www.math.ucla.edu/~cmarshak/GRE1.pdf.
One challenge for students preparing for the GRE mathematics subject test is the lack of resources: two
books, a few practice tests, and a some websites from previous workshops (many of which simply recycle
questions from the books and practice tests). It’s not enough, in my opinion. But as far as what’s best, I
recommend Cracking the GRE Mathematics Subject Test; the problems are a bit easier than what would be
ideal, but the questions are similar to those on the GRE mathematics subject test. I also recommend the
material from my friend’s 2012 UCLA GRE Workshop; you can view the practice problems at http://www.
math.ucla.edu/~cmarshak/GREWorkshop.html. The REA book GRE Mathematics is good too mainly due
to its difficulty level, but the authors weren’t able to emulate the actual GRE test as closely as the other
sources I cited.
My apologies for any errors. Alas, one of the problems with free documents is that it’s not feasible to hire
editors. And I must admit that I’m not the most careful writer. I welcome your help: If you find errors or have
questions feel free to email me at charles.tutoring@gmail.com. Feedback is greatly appreciated. The most
up-to-date version of this document can be found at http://www.rambotutoring.com/GR0568-solutions.
pdf.
To take care of a bit of shop work: If you find these solutions helpful, please consider purchasing my
solutions to GRE mathematics subject test 9768, which are on sale at http://www.rambotutoring.com/
booklets.php. I have a paperback booklet out that contains both sets of solutions, which is titled GRE
Mathematics Subject Test GR0568 and GR9768 Solutions: 1st edition. You can buy it at createspace.com
for $9.99. For details about my tutoring business check out my website http://www.rambotutoring.com.
I tutor throughout north San Diego county. Also, for search engine optimization purposes, a direct link to
http://www.rambotutoring.com would be appreciated, if you distribute this document.
Good luck on the GRE!
Charles Rambo,
Escondido, California
May 2013
Remarks for the 2nd edition
Hello everybody! I was so happy with the reception to my GR0568 solutions that decided to write solutions
to the GR9768 test too, which are currently on sale. But a funny thing happen: I found that the GR9768
solutions ended up substantially better than my GR0568 solutions. The second edition of this document is
meant to even things out.
I guess the most obvious changes are the title page and the glossary. Figuring out what should and shouldn’t
go in the glossary was tough, but I hope most of you will find that it contains all the information you need.
I chose to leave a lot of the theorems within the main text, especially the ones I felt are less well known to
third and fourth year math majors. For example, theorems from College Algebra were included.
There were a lot of other small changes too. I added new material to problems 56 and 61, and cleaned up
the solution to problem 57. I also went through the main text and made changes as needed; almost all the
questions ended up receiving minor rewrites. Lastly, many of the graphs were redone. Isn’t Gnuplot great?
I hope you enjoy the solutions and find them helpful. Thanks to everybody that emailed me. You guys
really helped improve the document, and getting the feedback is half the joy of making these things.
Charles Rambo,
Escondido, California
April 2014
GRE mathematics subject test GR0568 solutions
Question 1.
In the xy-plane, the curve with parametric equation x = cos t and y = sin t, 0 ≤ t ≤ π, has length
(A) 3
(B) π (C) 3π (D)
π
3
(E)
2
2
Solution. The arc length from t = a to t = b, where x and y are functions of a parameter t, is
Z
b
s
s=
a
dx
dt
2
+
dy
dt
2
dt.
Let’s find the pieces
x = cos t implies
dx
= − sin t
dt
and
y = sin t implies
dy
= cos t.
dt
It follows that
Z
s=
π
r
− sin t
2
2
+ cos t dt
Z0 π p
=
sin2 t + cos2 t dt
Z0 π
=
dt
0
= π.
Question 2.
Which of the following is an equation of the line tangent to the graph of y = x + et at x = 0?
(A) y = x (B) y = x + 1 (C) y = x + 2 (D) y = 2x
(E) y = 2x + 1
Solution. To find a tangent line, we need a point on the graph of y and the slope at that point. First, we
find the point. When x = 0, y = 0 + e0 = 1. Hence, the graph contains (0, 1). Next, we find the slope. The
derivative y 0 = 1 + ex gives the slope of the curve at x. It follows that when x = 0 the slope of our curve
is y 0 = 1+e0 = 2. Thus, the equation of the tangent line is y −1 = 2(x−1). Simplifying yields y = 2x+1.
1
Question 3.
If V and W are 2-dimensional subspaces in R4 , what are the possible dimensions of the subspace V ∩ W ?
(A) I only (B) 2 only (C) 0 and 1 only
(D) 0, 1, and 2 only (E) 0,1, 2, 3, and 4
Solution. Recall that the dimension of a vector space (or subspace) is defined to be the minimum number
of elements that span the space, i.e. the number elements in a basis of the vector space. We can extend a
basis of V ∩ W to a basis for either V or W , so the set of basis elements of V ∩ W must be a subset of a
basis of both V and W . So V ∩ W can have at most 2 basis elements (this case corresponds to V = W ) and
a minimum of 0 elements (corresponding to the case where V and W are disjoint except for the zero vector).
Of course, 1 basis element is also a possibility.
Question 4.
Let k be the number of real solutions of the equation ex + x − 2 = 0 in the interval [0, 1], and let n be the
number of real solutions that are not in [0, 1]. Which of the following is true?
(A) k = 0 and n = 1
(B) k = 1 and n = 0 (C) k = n = 1 (D) k > 1 (E) n > 1
Solution.
y
ex
x
−x + 2
We break the expression ex − (−x + 2) into two parts −x + 2 and ex , and consider where they intersect.
Using basic graphing techniques, it’s clear that there is exactly one point of intersection. Hence, ex + x − 2
only has one zero.
From here onward, it’s just a matter of finding where our zero lies. We will show our zero is in [0, 1] using
the Intermediate value theorem. Let f (x) := ex + x − 2 = ex − (−x + 2). The function f is real-valued
and continuous. It’s clear that f (0) = −1 and f (1) = e − 1 ≈ 1.7. So there must be a c in [0, 1] such that
f (c) = 0.
2
Question 5.
Suppose b is a real number and f (x) = 3x2 + bx + 12 defines a function on the real line, part of which is
graphed above. Then f (5) =
(A) 15
(B) 27 (C) 67 (D) 72 (E) 87
Solution. Since the coefficient in front of x2 is 3 and the vertex of f is (2, 0), it follows that f (x) = 3(x − 2)2 .
If you’re not familiar with vertex form, then consider the following argument. It’s clear f (2) = 2b + 24 = 0,
which implies b = −12. Either way, it follows that f (5) = 3(5 − 2)2 = 3(9) = 27.
Question 6.
Which of the following circles has the greatest number of points of intersection with the parabola x2 = y + 4.
(A) x2 + y 2 = 1 (B) x2 + y 2 = 2
(C) x2 + y 2 = 9 (D) x2 + y 2 = 16 (E) x2 + y 2 = 25
Solution.
3
By drawing a picture, we see there are at most 4 places where a circle centered at (0, 0) will intersect the
parabola. If we want all 4 intersections, the radius of our circle must be less than 4; if it’s equal to 4, then
it intersects the parabola at 3 points and if its radius is greater than 4 it intersects the parabola at 2 places.
Furthermore, the radius must be greater than or equal to 2, because the circle doesn’t intersect the parabola
at all if its radius is less than 2. Answer (C) is the only equation satisfying all of our criteria because its
radius is 3.
Question
7.
Z 3
|x + 1|dx =
−3
(A) 0 (B) 5 (C) 10 (D) 15 (E) 20
(
Solution. It’s clear |x + 1| =
Z
x + 1,
if x ≥ −1
. So,
−(x + 1), if x < −1
3
Z
3
|x + 1| dx =
−1
Z
−1
−(x + 1) dx
x + 1 dx +
−1
Z 3
−3
Z −1
x + 1 dx −
=
x + 1 dx
−1
−3
3
x2
+ x −
=
2
−1
9
= +3−
2
x2
+x
2
!
1
−1
2
= 10.
4
−
!−1
−3
!
1
−1 +
2
!
9
−3
2
Question 8.
What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1
and the other two vertices on the circle.
√
√
1
1+ 2
(A)
(B) 1 (C) 2 (D) π (E)
2
4
Solution.
y
h
θ
x
Without loss of generality, we can assume one side of our triangle on the positive x-axis. Let h and θ be
as shown in the picture. Due to symmetry, we need only consider values of θ between 0 and 180◦ . In this
interval, it’s clear h = sin θ.
From Geometry, we know the area A of our triangle is
1
A= ·1·h
2
sin θ
=
.
2
The maximum value of sin θ, for θ between 0 and 180◦ , is 1, so the maximum value of A must be 1/2.
Question 9.
Z
1
J=
p
1 − x4 dx
p
1 + x4 dx
p
1 − x8 dx
0
Z
1
K=
0
Z
L=
1
0
Which of the following is true for the definite integrals shown above.
(A) J < L < 1 < K (B) J < L < K < 1 (C) L < J < 1 < K (D) L < J < K < 1 (E) L < 1 < J < K
√
√
√
Solution. Because x8 < x4 when 0 < x < 1, it’s clear 1 − x4 ≤ 1 − x8 ≤ 1 ≤ 1 + x4 for 0 ≤ x ≤ 1,
where the inequality is strict except at x = 0 and 1. Therefore, due to basic integration properties,
Z 1p
Z 1p
Z 1p
1 − x4 dx <
1 − x8 dx < 1 <
1 + x4 dx.
0
0
0
5
Question 10.
Let g be the function whose derivative g 0 is continuous and has the graph shown above. Which of the
following values of g is largest?
(A) g(1)
(B) g(2) (C) g(3) (D) g(4) (E) g(5)
Solution. This is a simple application of the first derivative test. Our critical numbers are 2 and 5. Let’s
make a table to see where g is increasing and decreasing:
Intervals
Test numbers
g 0 (x)
g Increasing/Decreasing
(−∞, 2)
1
g 0 (1) > 0
Increasing
(2, 5)
3
g 0 (3) < 0
Decreasing
(5, ∞)
6
g 0 (6) > 0
Increasing
Let’s go through the ones that are false first. The number g(1) is nothing special, since g keeps increasing
until 2. Neither g(3) nor g(4) is a relative extreme, since g is decreasing between 2 and 5. And g(5) is a
relative minimum.
The first derivative test tells us that the value g(2) is a relative maxima. Hence, it must be the largest by
the process of elimination.
Question 11.
√
Of the following, which is the best approximation of 1.5(266)3/2 ?
(A) 1,000 (B) 2,700 (C) 3,200 (D) 4,100
(E) 5,300
Solution. It’s not too hard to see that
√ 1.5 · 266 = 399, which is approximately equal to the perfect square
400. Our strategy will be to modify 1.5(266)3/2 so that only 399 is within the radical. It’s clear
6
√
1.5(266)3/2 =
√
1.5 · 2663
√
= 266 1.5 · 266
√
= 266 399
√
≈ 266 400
= 266 · 20
= 5 320.
This would be a good time to select (E) during testing. But we’re not being timed (now), so let’s see how much
√
error we’re talking about.
From Calculus, we know f (x + dx) − f (x) ≈ df = f 0 (x)dx. Define f (x) := 266 x.
√
Then f 0 (400) = 133/ 400 = 6.65 and dx = 399 − 400 = −1. It follows that f (399) − f (400) ≈ −6.65, which
implies f (399) ≈ 5 320 − 6.65 = 5 313.35. Using a calculator, we see that f (399) = 5 313.34 . . ..
Question 12.
Let A be a 2 × 2 matrix for which there is a constant k such that the sum of the entries in each row and
each column is k. Which of the following must be an eigenvector of A?
1
I.
0
0
II.
1
1
III.
1
(A) I only (B) II only
(C) III only (D) I and II only (E) I, II, and III
a b
1
a
0
b
Solution. It’s clear A is of the form
, where a + b = k. Thus A
=
,A
=
, and
b a
0
b
1
a
1
a+b
k
1
1
A
=
=
=k
. So
is the only eigenvector listed.
1
a+b
k
1
1
Question 13.
A total of x feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible
area of the yard, in terms of x?
(A)
x2
9
(B)
x2
8
(C)
x2
4
(D) x2 (E) 2x2
Solution. Let the area in the yard be A := `w, where ` is the length and w is the width of the yard. By hypothesis, `+2w = x. Using substitution, it follows that A = (x−2w)w = xw −2w2 . Hence, dA/dw = x−4w.
Letting dA/dw = 0 and solving for w yields w = x/4. Because this is a parabola with a negative coefficient
in front of the w2 term, this critical number maximizes A. Thus, A(x/4) = (x − 2x/4)x/4 = x2 /8 is the
maximum area.
Question 14.
What is the units digit in the standard decimal expansion of the number 725 ?
7
(A) 1 (B) 3 (C) 5
(D) 7 (E) 9
Solution. This problem is equivalent to finding
725
(mod 10).
All that’s left is an easy computation:
725 ≡ 724 · 7
≡ 4912 · 7
≡ 912 · 7
≡ 816 · 7
≡ 16 · 7
≡7
(mod 10).
Question 15.
Let f be a continuous real-valued function defined on the closed interval [−2, 3]. Which of the following is
NOT necessarily true?
(A) f is bounded
R3
(B) −2 f (t) dt exists
(C) for each c between f (−2) and f (3), there is an x in [−2, 3] such that f (x) = c
R3
(D) there is an M in f ([−2, 3]) such that −2 f (t) dt = 5M
f (h) − f (0)
exists
(E) lim
h→0
h
Solution. Let’s go through the ones that are not necessarily true first. It’s not (A), because [−2, 3] is
compact and its image under a continuous map is compact, and by the Heine-Borel theorem every compact
space is bounded. Since f in continuous over the interval [−2, 3], the integral must exist, that means it
can’t be (B). Choice (C) is the Intermediate value theorem, and (D) is the integral form of the mean value
theorem.
By the process of elimination, it must be (E). And we can be confident about our choice, because
lim
h→0
f (h) − f (0)
=: f 0 (0),
h
which need not exist because differentiability is a stronger property than continuity. Consider f (x) = |x|; f
isn’t differentiable at 0, though it is continuous and real-valued on [−2, 3] .
8
Question 16.
What is the volume of the solid formed by revolving about the x-axis the region in the first quadrant of the
1
xy-plane bounded by the coordinate axes and the graph of the equation y = √
?
1 + x2
π
π2
π2
(D)
(A)
(B) π (C)
(E) ∞
2
4
2
Solution.
y
x
dx
√
2
The first step is to find a formula for an infinitesimal chunk volume
√ dV . At the point (x, 1/ 1 + x ), the
cross section perpendicular to the xy-plane is a disk of radius 1/ 1 + x2 , so its area must be
2
1
π
π √
=
.
2
1 + x2
1+x
It follows that
π
dx,
1 + x2
where dx is an infinitesimal horizontal length parallel to the x-axis. The volume
R of the generated solid is
simply the infinite sum of these infinitesimal chunks dV for all x ≥ 0 (i.e. V = x≥0 dV ). Hence,
Z ∞
π
V =
dx
1
+
x2
0
∞
= πArctan x
0
= π lim Arctan t − Arctan 0
t→∞
π
=π
−0
2
π2
.
=
2
A list of antiderivatives is contained in the glossary.
dV =
Question 17.
How many real zeros does the polynomial 2x5 + 8x − 7 have?
(A) None
(B) One (C) Two (D) Three (E) Five
Solution. We use Descartes’s rule of signs, which says:
9
Suppose f (x) = an xn + an−1 xn−1 + . . . + a1 x + a0 . Then the number of positive zeros of f is
equal to the number of sign changes of f (x) or is an even number less. Furthermore, the number
of negative zeros of f is equal to the number of sign changes of f (−x) or is an even number less.
Since there is one sign change in 2x5 + 8x − 7, Descartes’s rule of signs implies 2x5 + 8x − 7 has one positive
zero. To find the negative zeros, consider 2(−x)5 + 8(−x) − 7 = −2x5 − 8x − 7. There are no sign changes
so there are no negative zeros. Thus 2x5 + 8x − 7 has one real zero.
Question 18.
Let V be the real vector space of all real 2 × 3 matrices, and let W be the real vector space of all real
4 × 1 column vectors. If T is a linear transformation from V onto W, what is the dimension of the subspace
{v ∈ V : T (v) = 0}?
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6
Solution. Recall that the nullity of T and the rank of T are defined to be the dimensions of the null space
and the image, respectively. Since {v ∈ V : T (v) = 0} is the definition of the null space, all we need to do is
collect the pieces and apply the rank nullity theorem. The dimension of V is 6, and the dimension of W is
4. Since T is onto, the dimension of the rank must be the same as the dimension of the codomain, i.e. the
rank must be 4. It follows that n + 4 = 6, where n is the dimension of the null space. Thus, n = 2.
Question 19.
Let f and g be twice-differentiable real-valued functions defined on R. If f 0 (x) > g 0 (x) for all x > 0, which
of the following inequalities must be true for all x > 0?
(A) f (x) > g(x)
(B) f 00 (x) > g 00 (x)
(C) f (x) − f (0) > g(x) − g(0)
(D) f 0 (x) − f 0 (0) > g 0 (x) − g 0 (0)
(E) f 00 (x) − f 00 (0) > g 00 (x) − g 00 (0)
Solution. Due to elementary integration properties,
Z x
Z
0
f (t) dt >
0
x
g 0 (t) dt
0
because f 0 (x) > g 0 (x) for x > 0. Thus, the Fundamental theorem of Calculus implies that
f (x) − f (0) > g(x) − g(0).
10
Question 20.
Let f be the function defined on the real line by
(
f (x) =
x
2,
x
3,
if x is rational
if x is irrational.
If D is the set of points of discontinuity of f , then D is the
(A) empty set
(B) set of rational numbers
(C) set of irrational numbers
(D) set of nonzero real numbers
(E) set of real numbers
Solution. Let {qn } be a sequence of rational numbers such that qn → y as n → ∞, and let {xn } be a sequence
of irrationals such that xn → y as n → ∞. If f is continuous at y, then f (qn ) → y/2 and f (xn ) → y/3. But
limx→y f (x) cannot equal two different values if f is continuous at y, so we need y/2 = y/3. This can only
occur when y = 0. It follows that f is discontinuous everywhere except 0.
Question 21.
Let P1 be the set of all primes, {2, 3, 5, 7, . . .}, and for each integer n, let Pn be the set of all prime multiples
of n, {2n, 3n, 5n, 7n, . . .}. Which of the following intersections is nonempty?
(A) P1 ∩ P23 (B) P7 ∩ P21
(C) P12 ∩ P20 (D) P20 ∩ P24 (E) P5 ∩ P25
Solution. The set P12 ∩ P20 = {60} is nonempty because 12 · 5 = 20 · 3.
Let’s spend a bit of time thinking about the general rule. Suppose Pm ∩ Pn is nonempty and m 6= n. Then
there must be primes p and q such that m · p = n · q. When this is the case, m and n share all but one prime
factor; m has one more factor of q than n, and n has one more factor of p than m. In our case, 12 has one
more factor of 3 than 20, and 20 has one more factor of 5 than 12.
Question 22.
Let C(R) be the collection of all continuous functions from R to R. Then C(R) is a real vector space with
pointwise addition and scalar multiplication defined by
(f + g)(x) = f (x) + g(x) and (rf )(x) = rf (x)
for all f, g in C(R). and for all r, x in R. Which of the following are subspaces of C(R).
I. {f : f is twice differentiable and f 00 (x) − 2f 0 (x) + 3f (x) = 0 for all x}
II. {g : g twice differentiable and g 00 (x) = 3g 0 (x) for all x}
III. {h : h is a twice differentiable and h00 (x) = h(x) + 1 for all x}
(A) I only
(B) I and II only (C) I and III only (D) II and III only (E) I, II, and III
Solution. It’s clear that I and II are subspaces, but III isn’t because it’s not closed under scalar addition
(among other reasons). Suppose h is in {h : h is a twice differentiable and h00 (x) = h(x) + 1 for all x}, and
let r be in R. Then h00 (x) = h(x) + 1, or h(x) = h0 (x) − 1. It follows that rh(x) = rh00 (x) + r 6= rh00 (x) + 1
for r 6= 1.
Let’s take the time to check the subspace axioms for I and II, since we’re not on the clock.
11
First, I:
Closed under addition: Suppose f1 and f2 are in {f : f is twice differentiable and f 00 (x)−2f 0 (x)+
3f (x) = 0 for all x}. Then f1 (x) = 13 [f100 (x) − 2f10 (x)] and f2 (x) = 13 [f20 (x) − 2f20 (x)]. So,
(f1 + f2 )(x) = f1 (x) + f2 (x)
1
1
= [f100 (x) − 2f10 (x)] + [f20 (x) − 2f20 (x)]
3
3
1
00
= [(f1 + f2 ) (x) − 2(f1 + f2 )(x)] ,
3
which is in our subspace.
Closed under scalar multiplication: Let r be in R and f be in {f : f is twice differentiable and f 00 (x)−
2f 0 (x) + 3f (x) = 0 for all x}. Then
(rf )(x) = rf (x)
1
= [rf 00 (x) − 2rf 0 (x)]
3
1
= [(rf )00 (x) − 2(rf )0 (x)] ,
3
which in in our subspace.
Contains the zero vector: Consider the zero function O(x) := 0. It’s clear O(x) =
and (f + O)(x) = (O + f )(x) = f (x).
1
3
[O00 (x) − 2O0 (x)] = 0,
Now, II:
Closed under addition: Suppose g1 and g2 are in {g : g twice differentiable and g 00 (x) = 3g 0 (x) for all x}.
Then
00
(g1 + g2 ) (x) = g100 (x) + g200 (x)
= 3g10 (x) + 3g20 (x)
0
= 3 (g1 + g2 ) .
Closed under scalar multiplication: Suppose g is in {g : g twice differentiable and g 00 (x) =
3g 0 (x) for all x}. Then (rg)00 (x) = rg 00 (x) = 3(rg)0 (x).
Contains the zero vector: It’s the same zero vector O(x) := 0 as I, and it’s just as easy to prove
O(x) is in {g : g twice differentiable and g 00 (x) = 3g 0 (x) for all x}.
Question 23.
For what value of b is the line y = 10x tangent to the curve y = ebx at some point in the xy-plane?
(A)
10
(B) 10 (C) 10e (D) e10 (E) e
e
Solution. Let’s relabel to avoid confusion. Define f (x) := 10x and g(x) := ebx . If f is tangent to g at
x = x0 , then f 0 (x0 ) = g 0 (x0 ), i.e. 10 = bebx0 . Solving for x0 yields x0 = log(10/b)/b. To be tangent, we also
12
need f (x0 ) = g(x0 ), so
f
log(10/b)
b
10 log(10/b)
b
log(10/b)
=g
b
=
= elog(10/b)
10
=
.
b
That is, 10 log(10/b)/b = 10/b which implies log(10/b) = 1. Using logarithm properties to change the equation to exponential form gives 10/b = e1 .Therefore b = 10/e.
Question 24.
Z
Let h be the function defined by h(x) =
x2
ex+t dt for all real numbers x. The h0 (1) =
0
(A) e − 1 (B) e2 (C) e2 − e (D) 2e2
(E) 3e2 − e
Solution. We could differentiate this now using the Fundamental theorem of Calculus, but let’s try the
direct approach:
x2
Z
ex+t dt
h(x) =
0
x2
= ex+t = ex
2
t=0
+x
x
−e .
Note that x is fixed with respect to t, so x behaves like a constant when integrating with respect to t.
Now we differentiate:
h0 (x) =
2
d
ex +x − ex
dx
2
= (2x + 1)ex
+x
− ex .
Thus, we conclude h0 (1) = (2 + 1)e1+1 − e = 3e2 − e.
Question 25.
Let {an }∞
n=1 be defined recursively be a1 = 1 and an+1 =
(A) (15)(31) (B) (30)(31) (C)
n+2
n
an for n ≥ 1. Then a30 is equal to
31
32
32!
(D)
(E)
29
30
30!2!
Solution. Let’s see what a30 looks like:
31 30 29 28
6 5 4 3
a30 =
·
·
·
· . . . · · · · · 1.
29 28 27 26
4 3 2 1
So the numerator in the n-th factor will cancel with the denominator in (n + 2)-th factor most of the time.
When won’t this happen? Well, the numerators in the twenty-ninth and thirtieth factors won’t cancel,
13
because there are no thirty-first and thirty-second factors. Furthermore, since there is no −1-th factor and
the rule doesn’t hold for the first factor, the denominators in the second and third factors won’t cancel either.
With our cancelation rules in mind, we see
a30 = 31 · 30 ·
1
= (15)(31).
2
Question 26.
Let f (x, y) = x2 − 2xy + y 3 for all real x and y. Which of the following is true?
(A) f has all its relative extrema on the line x = y
(B) f
(C) f
(D) f
(E) f
has all of its relative extrema on the parabola x = y 2
has a relative minimum at (0, 0) has an absolute minimum at 32 , 23
has an absolute minimum at (1, 1)
Solution. Recall the second derivatives test from third semester Calculus:
Suppose that the function f : R2 → R has continuous second order partial derivatives in some
E ⊆ R2 . Suppose the point (a, b) in E is a critical point, i.e. fx (a, b) = 0 and fy (a, b) = 0. Define
f (x, y) fxy (x, y)
2
Hf (x, y) := det xx
= fxx (x, y)fyy (x, y) − [fxy (x, y)] .
fyx (x, y) fyy (x, y)
Then
• if fxx (a, b) > 0 and Hf (a, b) > 0, then f (a, b) is a relative minimum;
• if fxx (a, b) < 0 and Hf (a, b) > 0, then f (a, b) is a relative maximum;
• if Hf (a, b) < 0, then (a, b) is a saddle point;
• if Hf (a, b) = 0, then the test gives no information.
Let’s find our critical points. The partial derivative with respect to x and y are
fx (x, y) = 2x − 2y and fy (x, y) = −2x + 3y 2 ,
respectively. Our critical numbers occur when our first order partial derivates are zero, so let 2x − 2y = 0
and −2x + 3y 2 = 0. The first equation implies that all extreme must lie on x = y so we’re ready to choose
(A), but let’s keep going.
We need the critical numbers. Using substitution and a bit of algebra on the equations 2x − 2y = 0 and
−2x + 3y 2 = 0, we conclude 3x2 − 2x = x(3x − 2) = 0. That means our critical numbers are (0, 0) and
(2/3, 2/3).
Next, we find Hf (x, y). The second order partial derivatives are:
fxx (x, y) = 2,
fxy (x, y) = fyx (x, y) = −2,
and fyy (x, y) = 6y.
It follows that Hf (x, y) = 12y − 4.
With the second derivatives test in mind, we examine our critical points.
• (0, 0): Hf (0, 0) = −4 > 0, so (0, 0) is a saddle point;
• (2/3, 2/3): fxx (2/3, 2/3) = 2 > 0 and Hf (2/3, 2/3) = 8−4 = 4 > 0, so f (2/3, 2/3) is relative minimum.
14
Obviously, theses results exclude (C), but could (2/3, 2/3) be an absolute minimum? No. Why? Because f
has no lower bound: fix x and you can see f (x, y) = x2 − 2xy + y 3 → −∞ as y → −∞.
Question 27.
Consider the two planes x + 3y − 2z = 7 and 2x + y − 3z = 0 in R3 . Which of the following sets is the
intersection of these planes?
(A) ∅
(B) {(0, 3, 1)}
(C) {(x, y, x) : x = t, y = 3t, z = 7 − 2t, t ∈ R}
(D) {(x, y, x) : x = 7t, y = 3 + t, z = 1 + 5t, t ∈ R}
(E) {(x, y, x) : x − 2y − z = −7}
Solution. We will use simultaneous equations to solve this problem. Since most of our options have simple
functions describing the x variable, we solve for x first.
−3(2x + y − 3z) = −6x − 3y + 9z
=0
So,
+
−6x
x
−5x
−3y
+3y
+9z
−2z
+7z
=0
=7
= 7,
or z = 1 + 5x/7. To remove the fraction, let x = 7t. This implies z = 1 + 5t. We were given that
2x + y − 3z = 0, so y = −2x + 3z = −2(7t) + 3(1 + 5t) = t + 3. Thus the solution is {(x, y, x) : x = 7t, y =
3 + t, z = 1 + 5t, t ∈ R}.
Question 28.
The figure above shows an undirected graph with six vertices. Enough edges are to be deleted from the
graph in order to leave a spanning tree, which is a connected subgraph having the same six vertices and no
cycles. How many edges must be deleted?
(A) One (B) Two (C) Three
(D) Four (E) Five
Solution. Let blue marks denote deletion of a side. The graph on the next page satisfies the necessary
criteria.
15
.
Question 29.
For all positive f and g of real variable, x, let ∼ be a relation defined by
f ∼ g if and only if lim
x→∞
f (x)
g(x)
Which of the following is NOT a consequence of f ∼ g?
(A) f 2 ∼ g 2 (B)
√
f∼
√
g
(C) ef ∼ eg (D) f + g ∼ 2g (E) g ∼ f
Solution. Let’s find a counterexample to ef ∼ eg . Consider f (x) = x and g(x) = x + 1. It’s clear
f (x)
= 1. But
lim
x→∞ g(x)
lim
x→∞
ex
ex+1
= lim
x→∞
1
e
1
e
6 1.
=
=
This is enough to move on, but we’re not being timed so let’s verify that the others work.
First, f 2 ∼ g 2 .
2
2
f (x)
f (x)
lim =
lim
2
x→∞ g(x)
x→∞
g(x)
= 12
=1
Next,
√
f∼
√
g. The argument is very similar.
16
s
p
f (x)
f (x)
=
lim p
lim
x→∞
x→∞ g(x)
g(x)
√
= 1
=1
We continue by considering f + g ∼ 2g.
lim
x→∞
f (x) + g(x)
1
f (x)
=
lim
+1
x→∞
2g(x)
2
g(x)
1
= (1 + 1)
2
=1
Onto g ∼ f .
lim
x→∞
1
g(x)
= lim f (x)
x→∞
f (x)
g(x)
1
=
limx→∞
f (x)
g(x)
1
1
= 1.
=
Question 30.
Let f be a function from a set X to a set Y . Consider the following statements.
1. P : For each x in X, there exists y in Y such that f (x) = y.
2. Q: For each y in Y , there exists x in X such that f (x) = y.
3. R: There exists x1 and x2 in X such that x1 6= x2 and f (x1 ) = f (x2 ).
The negation of the statement “f is one-to-one and onto Y ” is
(A) P or not R (B) R or not P
(C) R or not Q (D) P and not R (E) R and not Q
Solution. The negation of “f is one-to-one and onto Y ” is “f is not one-to-one or f is not onto Y ” (remember that “or” is inclusive). Statement R is “f is not one-to-one” and Q means f is onto, so not Q must
mean f is not onto. We can conclude that R or not Q is the correct answer.
17
Question 31.
Which of the following most closely represents the graph of a solution to the differential equation
dy
= 1+y 4 ?
dx
It’s easy to see that dy/dx → ∞ as |y| → ∞, which narrows our selection down to (A) and (B). Since
dy/dx > 0, we exclude choice (B) because the derivative of that function is about 0 when its graph intersects
the y-axis.
18
Question 32.
Suppose that two binary operations, denoted by ⊕ and , are defined on a nonempty set S, and that the
following conditions are satisfied for all x, y, and z in S:
(1) x ⊕ y and x y are in S.
(2) x ⊕ (y ⊕ z) = (x ⊕ y) ⊕ z and x (y z) = (x y) z
(3) x ⊕ y = y ⊕ x
Also, for each x in S and for each positive integer n, the elements nx and xn are defined recursively as
follows:
1x = x1 = x and
if kx and xk have been defined, then (k + 1)x = kx ⊕ x and xk+1 = xk x.
Which of the following must be true?
I. (x y)n = xn y n for all x and y in S and for each positive integer n.
II. n(x ⊕ y) = nx ⊕ ny for all x and y in S and for each positive integer n.
III xm xn = xm+n for each x in S and for all positive integers m and n.
(A) I only (B) II only (C) III only
(D) II and III only (E) I, II, and III
Solution. We can immediately exclude property I, because it fails whenever is not commutative.
Property II follows. Consider
n(x ⊕ y) = (x ⊕ y) ⊕ (x ⊕ y) ⊕ · · · ⊕ (x ⊕ y)
{z
}
|
n times
= (x ⊕ x ⊕ · · · ⊕ x) ⊕ (y ⊕ y · · · ⊕ y)
|
{z
} |
{z
}
n times
n times
= nx ⊕ ny.
Property III follows due to a similar argument.
xn xm = (x x · · · x) (x x · · · x)
|
{z
} |
{z
}
n times
= (x x · · · x)
|
{z
}
n+m times
= xn+m .
19
m times
Question 33.
The Euclidean algorithm is used to find the greatest common divisor (gcd) of two positive integers a and b.
input(A)
input(B)
while b>0
begin
r := a mod b
a := b
b := r
end
gcd := a
output (gcd)
When the algorithm is used to find the greatest common divisor of a = 273 and b = 110, which of the
following is the sequence of computed values for r?
(A) 2, 26, 1, 0 (B) 2, 53, 1, 0 (C) 53, 2, 1, 0
(D) 53, 4, 1, 0 (E) 53, 5, 1, 0
Solution. Let’s go through the process. We plug in a = 273 and b = 110, and our first r = 273 mod 110
= 53. For the next step a = 110 and b = 53, so r = 110 mod 53 = 4. Then a = 53, b = 4, so r = 1.
Then a = 4, b = 1, so r = 0. Thus, our sequence of r’s is 53, 4, 1, 0.
Question 34.
The minimum distance between any point on the sphere (x − 2)2 + (y − 1)2 + (z − 3)3 = 1 and any point on
the sphere (x + 3)2 + (y − 2)2 + (z − 4)2 = 4 is
√
(A) 0 (B) 4 (C)
√
27 (D) 2( 2 + 1)
√
(E) 3( 3 − 1)
Solution. Since the line containing our points also contains the centers of the spheres, all we need to do is
calculate the distance between the
The distance between
pcenters of the spheres, and subtract
√ the two radii. √
the centers of the spheres is d = (−3 − 2)2 + (2 − 1)2 + (4 − 3)2 = 25 + 1 + 1 = 3 3. The radius of the
first sphere is 1, and
√ the second’s is√2. It follows that the distance, between the points closest to each other
on the spheres is 3 3 − 1 − 2 = 3( 3 − 1).
Question 35.
At a banquet, 9 women and 6 men are to be seated in a row of 15 chairs. If the entire seating arrangement
is to be chosen at random, what is the probability that all of the men will be seated next to each other in 6
consecutive positions?
(A)
1
(B)
15
6
6!
(C)
15
6
6!10!
10!
(D)
15!
15!
(E)
6!10!
15!
Solution. We will use the Fundamental counting principle extensively. First, the total number of permutations of men and women is 15!. To find the number of permutations where all the men are seated next to
each other, consider the collection of men to be one unit, which gives us 10! permutations of the women and
the “men-unit”. Within the “men-unit”, there are 6! ways to seat the group of men. It follows that there
are a total of 10!6! ways to seat the group if all the men sit together. Thus, the probability that all the men
will sit together is 10!6!/15!.
20
Question 36.
Let M be a 5 × 5 real matrix. Exactly four of the following five conditions on M are equivalent to each
other. Which of the five conditions is equivalent to NONE of the other four?
(A) For any two distinct column vectors u and v of M , and set {u, v} is linearly independent
(B) The homogeneous system M x = 0 has only the trivial solution
(C) The system of equations M x = b has a unique solution for each real 5 × 1 column vector b
(D) The determinant of M is nonzero
(E) There exists a 5 × 5 real matrix N such that N M is the 5 × 5 identity matrix.
Solution. All but (A) are claims equivalent to saying the columns of M are linearly independent. Recall
M having linearly independent columns implies the only solution to M x = 0 is x = 0 (B), which implies
every equation of the form M x = b has only one solution (C), which implies that M is invertible (E), which
implies that the determinant of M is nonzero (D).
To see why (A) isn’t equivalent to the rest, consider

1 0
0 1

M =
0 0
0 0
0 0
0
0
1
0
0
0
0
0
1
0

1
1

1
.
1
0
Clearly, the columns aren’t linearly independent, though they are pairwise independent.
Question 37.
In the complex z-plane, the set of points satisfying the equation z 2 = |z|2 is a
(A) pair of points (B) circle (C) half-line
(D) line (E) union of infinitely many different lines
Solution. Let z = x + iy, where x and y are real numbers. Then z 2 = (x + iy)2 = x2 − y 2 + 2ixy and
|z|2 = x2 + y 2 . It’s clear this implies y 2 = ixy, but a real number can’t equal an imaginary one, so y = 0.
Thus the equation must describe a line.
Question 38.
Let A and B be an nonempty subset of R and let f : A → B be a function. If C ⊂ A and D ⊂ B, which of
the following must be true?
(A) C ⊆ f −1 (f (C))
(B) D ⊆ f f −1 (D)
(C) f −1 (f (C)) ⊆ C
(D) f −1 (f (C)) = f f −1 (D)
(E) f f −1 (D)
Solution. The definition of the set f −1 (Y ) is {x ∈ A : f (x) ∈ Y }, where Y ⊆ B. So it’s clear that
C ⊆ f −1 (f (C)) because x in C implies f (x) in f (C) so x in f −1 (f (C)).
21
Question 39.
In the figure above, as r and s increase, the length of the third side of the triangle remains 1 and the measure
of the obtuse angle remains 110◦ . What is s→∞
lim (s − r).
r→∞
(A) 0
(B) a positive number less than 1 (C) 1 (D) a finite number greater than 1 (E) ∞
Solution. We use the Law of cosines. The identity states
c2 = a2 + b2 − 2ab cos C,
where a, b, and c are sides of a triangle and C is the angle opposite c.
In this situation the identity implies s2 = 1+r2 −2r cos 110◦ , and taking square roots yields s =
lim (s − r) = lim
s→∞
r→∞
r→∞
p
√
1 + r2 − 2r cos 110◦ .
1 + r2 − 2r cos 110◦ − r
1 + r2 − 2r cos 110◦ − r2
= lim √
r→∞
1 + r2 − 2r cos 110◦ + r
1 − 2r cos 110◦
= lim √
r→∞
1 + r2 − 2r cos 110◦ + r
1
− 2 cos 110◦
= lim q r
r→∞
1
2 cos 110◦
+1
r2 + 1 −
r
= − cos 110◦
= cos 70◦ .
In conclusion, 0 < cos 70◦ < cos 60◦ = 1/2 < 1, so the solution must be (B).
Question 40.
For which of the following rings is it possible for the product of two nonzero elements to be zero?
(A) the ring of complex numbers
(B) the ring of integers modulo 11
(C) the ring of continuous real-valued functions on [0, 1]
√
(D) the ring {a + b 2 : a and bare rational numbers}
(E) the ring of polynomials in x with real coefficients
Solution. Consider
(
0,
if x ≥ 1/2
f (x) :=
x − 1/2, if x < 1/2
(
and
22
g(x) :=
x − 1/2, if x ≥ 1/2
0,
if x < 1/2.
It’s clear that both of these functions are continuous on [0, 1] and real valued. It’s also clear (f g)(x) = 0.
Question 41.
Let C beI the circle x2 + y 2 = 1 oriented counterclockwise in the xy-plane. What is the value of the line
integral (2x − y) dx + (x + 3y) dy?
(A) 0 (B) 1 (C)
π
(D) π
2
(E) 2π
Solution. Green’s theorem says that
I
(2x − y) dx + (x + 3y) dy =
C
ZZ
∂
∂
(x + 3y) −
(2x − y) dA
∂x
∂y
{(x,y): x2 +y 2 <1}
ZZ
=
1 + 1 dA
{(x,y):
x2 +y 2 <1}
= 2A
= 2π.
Question 42.
Suppose X is a discrete random variable on the set of positive integers such that for each positive integer n,
the probability that X = n is 21n . If Y is a random variable with the same probability distribution X and
Y are independent, what is the probability that the value of at least one of the variables X and Y is greater
than 3?
1
1
3
4
15
(A)
(C)
(D)
(E)
(B)
64
64
4
8
9
Solution. This is simply an application of a few probability properties. Consider the following argument.
P (X > 3 or Y > 3) = 1 − P (X ≤ 3 and Y ≤ 3)
= 1 − P (X ≤ 3) · P (X ≤ 3)
2
= 1 − P (X ≤ 3)
1 1 1 2
=1−
+ +
2 4 8
49
=1−
64
15
=
.
64
23
Question 43.
2πi
If z = e 5 , then 1 + z 2 + z 3 + 5z 4 + 4z 5 + 4z 6 + 4z 7 + 4z 8 + 5z 9 =
(A) 0 (B) 4e
3πi
5
(C) 5e
4πi
5
(D) −4e
2πi
5
(E) −5e
3πi
5
Solution. Recall Euler’s formula:
eiθ = cos θ + i sin θ.
In particular, we need that e2πi = 1 and eπi = −1.
Hence, z 5 = 1, which implies z 5 − 1 = (z − 1)(z 4 + z 3 + z 2 + z + 1) = 0, since z 6= 1 we can divide both
sides by z − 1, which gives us z 4 + z 3 + z 2 + z + 1 = 0. From here, it’s only a matter of substitution and
factorization.
1 + z 2 + z 3 + 5z 4 + 4z 5 + 4z 6 + 4z 7 + 4z 8 + 5z 9 = (1 + z 2 + z 3 + z 4 ) + 4z 4 (1 + z + z 2 + z 3 + z 4 ) + 5z 9
= 5z 9
= 5 e2πi/5
9
= 5e2πi e8πi/5
= 5e8πi/5
= 5eπi e3πi/5
= 5(−1)e3πi/5
= −5e3π/5 .
Question 44.
A fair coin is to be tossed 100 times, with each toss resulting in a head or a tail. If H is the total number
of heads and T is the total number of tails, which of the following events has the greatest probability?
(A) H = 50 (B) T ≥ 60 (C) 51 ≤ H ≤ 55
(D) H ≤ 48 and T ≥ 48 (E) H ≤ 5 or H ≥ 95
Solution. The probability distribution for the number of heads occurring on a fair coin is a binomial
distribution. In general, for a binomial distribution:
• The expected value is µ = np
• The standard deviation is σ =
p
np(1 − p),
where n is the number of trials and p is the probability of the affirmative case. It’s also helpful to note that
for large n, the normal distribution is a good approximation of a binomial distribution.
Onto
our problem. We were given that n = 100 and p = 1/2. So µ = 100(1/2) = 50 and σ =
p
100(1/2)(1/2) = 5. Since the probability of obtaining any specific number of heads or tails is quite
low, we exclude (A). As mentioned above, the normal distribution is a good approximation for the binomial
distribution, so the probability T ≥ 60 or H ≤ 40 is low, because it’s two standard deviations from the
expected result of 50. As such, exclude (B). Exclude (E) because those results are even further away from
the 50. We’ve narrowed it down to (C) and (D). Choice (D) is equivalent to stating 48 ≤ H ≤ 52. So the
events in (D) are a bit closer to the expected value of 50 than those in 51 ≤ H ≤ 55. Therefore we conclude
the solution is (D).
24
Question 45.
A circular region is divided by 5 radii into sectors as shown above. Twenty-one points are chosen in the
circular region, none of which is on any of the 5 radii. Which of the following must be true?
I. Some sector contains at least 5 of the points.
II. Some sector contains at most 3 of the points.
III. Some pair of adjacent sectors contains a total of at least 9 of the points.
(A) I only (B) III only (C) I and II only
(D) I and III only (E) I, II, and III
Solution. This is a simple application of pigeonhole principle. However, we won’t reference it directly in
our argument.
Statement I is true. Since there will always be twenty-one points, less in one sector implies more in another.
As such, to minimize the number of points in the sector with the most, we distribute the points as homogeneously as possible. So, we would want to put 21/5 = 4.2 points in each sector, but this is impossible
since we can’t put 0.2 points in a sector. Thus the most homogenous distribution would have 5 points in
one sector.
Statement II is a lie. The situation described in option I proves that II isn’t necessarily true, since we could
have 4 points in four sectors and 5 in the remaining sector.
Statement III is a legitimate proposition. Consider new sectors created by combining the adjacent sectors
in the picture above. Each sector is adjacent to two others, so we double counted the old sectors. Therefore,
we’ve also double counted the number of points, so our new sectors have a combined total of 42 points.
There are a total of 5 new sectors. To minimize the number of points in the new sector with the most, we
want to distribute the points as homogeneously as possible. That means we want to put 42/5 = 8.4 points
in each new sector. But the most we can put in any new sector is 8, so one new sector must contain an
additional point.
25
Question 46.
Let G be the group of complex numbers {1, i, −1, −i} under multiplication. Which of the following statements
are true about the homomorphisms of G into itself?
I. z 7→ z̄ defines one such homomorphism, where z̄ denotes the complex conjugate of z.
II. z 7→ z 2 defines one such homomorphism.
III. For every such homomorphism, there is an integer k such that the homomorphism has the form z 7→ z k .
(A) None (B) II only (C) I and II only (D) II and III only
(E) I, II, and III
Solution. We need the following identity:
Let z be a complex number, and |z| the modulus of z. Then z = |z|eiθ = |z| (cos θ + i sin θ), for
some θ.
Since the modulus of each element in G is 1, the above identity simplifies to z = eiθ .
Onto our problem. It’s clear II follows from III. Choice III also implies I, because z 7→ z̄ is the same as
z 7→ z −1 (since z −1 = e−iθ = cos(−θ) + i sin(−θ) = cos θ − i sin θ). So all we need to do is prove III. The
exponent rules for multiplication prove that z 7→ z k will be a homomorphism. Because i generates the group,
all that’s left to show is that every map takes i to ik for some integer k. Since i = eiπ/2 , it’s not too tough
to see that i 7→ 1 = i4 , i 7→ −1 = i2 , i 7→ −i = i3 , and i 7→ i = i1 .
Question 47.
Let F be a constant unit force that is parallel to the vector (−1, 0, 1) in xyz-space. What is the fork done
by F on a particle that moves along the path by (t, t2 , t3 ) between time t = 0 and time t = 1?
(A) −
1
1
(B) − √
4
4 2
(C) 0 (D)
√
√
2 (E) 3 2
Solution. We need the formula for work done.
Let C := {γ(t) : a ≤ t ≤ b}, where γ : R → R3 is a differentiable in each slot. Then the work
done by a vector field F over C is
Z
b
Z
F · γ 0 (t) dt.
F · dγ =
W =
C
a
In our case, the formula above implies
Z
(−1, 0, 1) · dγ
W =
{(t,t2 ,t3 ):
Z
0≤t≤1}
1
(−1, 0, 1) · (1, 2t, 3t2 ) dt
=
0
Z
=
1
−1 + 3t2 dt
0
1
= −t + t3 0
= −1 + 1 − (0)
= 0.
26
Question 48.
Consider the theorem: If f and f 0 are both strictly increasing real-valued functions on the interval (0, ∞),
then lim f (x) = ∞. The following argument is suggested as a proof of this theorem.
x→∞
(1) By the Mean Value Theorem, there is a c1 in the interval (1, 2) such that
f 0 (c1 ) =
f (2) − f (1)
= f (2) − f (1) > 0.
2−1
(2) For each x > 2, there is a cx in (2, x) such that
(3) For each x > 2,
f (x) − f (2)
= f 0 (cx ).
x−2
f (x) − f (2)
= f 0 (cx ) > f 0 (c1 ) since f 0 is strictly increasing.
x−2
(4) For each x > 2, f (x) > f (2) + (x − 2)f 0 (c1 ).
(5) lim f (x) = ∞
x→∞
Which of the following is true?
(A) The argument is valid.
(B) The argument is not valid since the hypotheses of the Mean Value Theorem are not satisfied in (1)
and (2).
(C) The argument is not valid since (3) is not valid.
(D) The argument is not valid since (4) cannot be deduced from the previous steps.
(E) The argument is not valid since (4) does not imply (5).
Inane ramblings. The argument is valid. There’s a moral to this question: Don’t assume that it can’t be
any particular answer until you’ve examined the math. According to the GRE booklet, 63% of test takers got
this question wrong. I assume because they thought (A) was the “sucker’s answer” and worked themselves
to death trying to find nonexistent errors.
Question 49.
Up to isomorphism, how many additive abelian groups G of order 16 have the property that x + x + x + x = 0
for each x in G?
(A) 0 (B) 1 (C) 2
(D) 3 (E) 5
Solution. We use the finite case of the Fundamental theorem of finitely generated abelian groups.
Let G be a finite abelian group of order m. Then it is isomorphic to an expression of the form
Zk1 × Zk2 × . . . × Zkn ,
where ki divides both ki+1 and m for all i = 1, 2, . . . , n − 1 and m = k1 · k2 · . . . · kn .
In our situation, we can’t have a ki greater than 4 because that would imply that the characteristic of G
would be greater than 4. Thus, the only groups with the property that x + x + x + x = 0 are
1. Z4 × Z4
2. Z2 × Z2 × Z4
27
3. Z2 × Z2 × Z2 × Z2 .
Question 50.
Let A be a real 2 × 2 matrix. Which of the following statements must be true?
I. All of the entries of A2 are nonnegative.
II. There determinant of A2 is nonnegative.
III. If A has two distinct eigenvalues, then A2 has two distinct eigenvalues.
(A) I only
(B) II only (C) III only (D) II and III only (E) I, II, and III
Solution. Statement I is a lie. Consider
A :=
0
1
−1
.
0
Then
A2 :=
0
.
−1
−1
0
Statement II is true, because det(A2 ) = det(A) · det(A) ≥ 0.
Statement III is a fib. What if A has eigenvalues are λ1 = 1 and λ2 = −1? Doesn’t that imply A2 ’s only
eigenvalue would be 1?
Question 51.
∞
Z
bxce−x dx =
If bxc denotes the greatest integer not exceeding x, then
0
(A)
e
e2 − 1
(B)
1
e−1
(C)
(D) 1 (E) +∞
e−1
e
Solution. It’s not too tough to see
Z
∞ Z
X
∞
bxce
−x
dx =
0
The next few steps are just simple math
∞ Z n+1
X
n=1
n=1
ne−x dx =
n
=
∞
X
n+1
ne−x dx.
n
−ne−n−1 + ne−n
n=1
∞
X
n(1 − e−1 )
en
n=1
= (1 − e−1 )
= (1 − e−1 )
∞
X
n
en
n=1
∞
X
n=1
We need to find a formula for
∞
X
nxn .
n=1
28
n e−1
n
.
A bit of cleverness is required here. But it’s only canned cleverness, so you may need to use some of the
tricks below for the GRE. First, recall the formula for an infinite geometric series
∞
X
1
=
xn ,
1 − x n=0
when |x| < 1. It follows that
d
1
1
=
dx 1 − x
(1 − x)2
!
∞
d X n
=
x
dx n=0
=
∞
X
nxn−1 .
n=1
Multiplying both sides by x yields
∞
X
x
=
nxn .
(1 − x)2
n=1
Notice that the left side of the equation above is what we’re looking for, where x = e−1 . Thus,
(1 − e−1 )
∞
X
n e−1
n
= (1 − e−1 )
n=1
=
e−1
(1 − e−1 )2
1
.
e−1
Question 52.
If A is a subset of the real line R and A contains each rational number, which of the following must be true?
(A) If A is open, then A = R.
(B) If A is closed, then A = R.
(C) If A is uncountable, then A = R.
(D) If A is uncountable, then A is open.
(E) If A is countable, then A is closed.
Solution. Option (B) is true. Here’s why: Recall that Q is dense in R, i.e. cl(Q) = R where cl denotes the
closure of a set. It follows that if A is closed cl(Q) = R ⊆ cl(A) = A ⊆ cl(R) = R. Thus A = R.
Let’s see why the others are lies. To see why (A) isn’t true, consider A = R \ {π}; the set A is open because
for each x in A, the open ball centered at x with radius |x − π|/2 is contained in A. To disprove (C), consider
√
A := Q ∪ (0, 1). The same counterexample disproves (D). Choice (E) is disproved by A = Q, because 2 is
a limit point.
29
Question 53.
What is the minimum value of the expression x + 4z as a function defined on R3 , subject to the constraint
x2 + y 2 + z 2 ≤ 2?
(A) 0 (B) -2
√
√
√
(C) − 34 (D) − 35 (E) −5 2
Solution. The minimum value of f (x, y, z) := x+4z occurs on the boundary of the solid sphere x2 +y 2 +z 2 ≤
2, i.e. the minimum value for f occurs when x2 + y 2 + z 2 = 2.
We need the method of Lagrange multipliers:
Suppose f (x, y, x) had first order partial derivatives, and the values x, y, and z satisfy the
constraint g(x, y, z) = k, where k is a constant. Then
fx (x, y, x) = λgx (x, y, z),
fy (x, y, z) = λgy (x, y, z),
and fz (x, y, z) = λgz (x, y, z)
for some λ in R.
Obviously, the function we wish to maximize is f (x, y, z) := x + 4z and our constraint is g(x, y, z) :=
x2 + y 2 + z 2 = 2. It follows that
1 = 2λx,
0 = 2λy,
and 4 = 2λz.
A bit of Algebra shows y = 0, and z = 4x. Substituting the right sides into the appropriate slots of g yields
g(x, 0, 4x) = x2 + (4x)2
= x2 + 16x2
= 17x2
= 2.
p
√
√
√
√
√
So, x = ± 2/17 = ± 34/17 and z = ±4 34/17, which√means f (± 34/17, 0, ±4 34/17) = ± 34. Thus,
the minimum value of f satisfying x2 + y 2 + z 2 = 2 is − 34.
30
Question 54.
The four shaded circles in Figure 1 above are congruent and each is tangent to the large circle and to two of
the other shaded circles. Figure 2 is the result of replacing each of the shaded circles in Figure 1 by a figure
that is geometrically similar to Figure 1. What is the ratio of the area of the shaded portion of Figure 2 to
the area of the shaded portion of Figure 1?
√ !2
2
2
1
1
4
2
√ (C)
√ (D)
√
√
(A) √ (B)
(E)
2 2
1+ 2
1+ 2
1+ 2
1+ 2
Solution. Here’s a zoomed in version of a shaded circle in figure 1.
y
r
√
r 2
r
x
r
Let’s find the ratio of the shaded area in figure 1 to the area of the big circle in figure
√ 1. Suppose
√ the big
circle has radius R and the
small
circle
has
radius
r.
Then
our
picture
implies
r
+
r
2
=
r(1
+
2) = R. It
√
follows that r = R/(1 + 2). Hence, the ratio of the 4 shaded circles in figure 1 to the area of the big circle
is
2
2
4π 1+R√2
2
√
=
.
πR2
1+ 2
Since the ratio of the big circle in figure 1 to the small shaded circles in figure 1 is the same as the ratio of
2
the shaded circles in figure 2 to shaded circles in figure 1, we conclude the answer is 1+2√2 .
31
Question 55.
For how many positive integers k does the ordinary decimal representation of the integer k! end in exactly
99 zeros?
(A) None (B) One (C) Four
(D) Five (E) Twenty-four
Solution. Zeros are added every time another factor of 10 appears (after perhaps multiplying some numbers
together). Since there are more even numbers than multiples of 5, it follows that new zeros are added to the
decimal representation of k! whenever k is a multiple of 5.
Let’s obtain some rules to help us complete this problem. The number 5! ends in one zero, 10! ends in two
zeros, and 15! ends in three. The pattern continues until we reach 25!, which ends in six zeros since 25 has
two factors of 5. It’s clear this generalizes to the following rule: When k has two factors of 5, two zeros
are added to the decimal representation of k!. Going from 124! to 125! adds three new zeros, since 125 has
three factors of 5. It’s clear this generalizes to the following: When k has three factors of 5, three zeros are
added to the decimal representation of k!. In general: When k has n factors of 5, n new zeros appear in the
decimal representation of k!.
From here, we consider values of k and use the principles outlined above. Suppose k = 400. Then
400! = 400 · 399 · . . . · 1 has 80 numbers which have 5 as a factor, 16 of which are multiples of 25, and
3 are multiples of 125. It follows that 400! will have 80 + 16 + 3 = 99 zeros. Clearly 401!, 402!, 403!, and 404!
will also have 99 zeros, since 401, 402, 403, and 404 have no new factors of five.
Question 56.
Which of the following does NOT define a metric on the set of all real numbers?
(
0 if x = y
(A) δ(x, y) =
2 if x 6= y
(B) ρ(x, y) = min{|x − y|, 1}
|x − y|
(C) σ(x, y) =
3
|x − y|
(D) τ (x, y) =
|x − y| + 1
(E) ω(x, y) = (x − y)2
Solution. For those unfamiliar with the definition of a metric, see the glossary at the end of this booklet.
Since r2 < |r| when |r| < 1, ω fails the triangle inequality. For example, say x = 1, y = 0, and z = 1/2.
Then ω(1, 0) = 1 and ω(1, 1/2) + ω(1/2, 0) = 1/4 + 1/4 = 1/2.
This is where we would move on during the test, but let’s make a few more remarks. Verifying that (A)–(C)
are metrics doesn’t add much insight. Instead, we’ll explain why we would want to examine τ and ω first
during the GRE:
(A) It looks like δ behaves like the discrete metric.
(B) It looks like ρ behaves like the Euclidian metric for values “close” together, and behaves like the discrete
metric elsewhere.
(C) It looks like σ is equivalent to the Euclidian metric.
Now let’s verify that τ is a metric. We know d(x, y) = |x − y| is a metric, and non-negativity and symmetry
32
of τ follow directly from this. Next, note that
f (x) :=
x
1+x
is monotonically increasing for all non-negative numbers. You can prove this, for example, by taking the
derivative. Hence, τ passes the triangle inequality criterion, because
τ (x, y) = f (|x − y|)
≤ f (|x − z| + |z − y|)
|x − z| + |z − y|
1 + |x − z| + |z − y|
|x − z|
|z − y|
=
+
1 + |x − z| + |z − y| 1 + |x − z| + |z − y|
|x − z|
|z − y|
≤
+
1 + |x − z| 1 + |z − y|
= τ (x, z) + τ (z, y),
=
for all z.
Question 57.
The set of real numbers x for which the series
∞
X
n=1
nn
n!x2n
converges is
(1 + x2n )
√
√
(A) {0} (B) {x : −1 < x < 1} (C) {x : −1 ≤ x ≤ 1} (D) {x : − e ≤ x ≤ e}
(E) R
Solution. It’s clear that the answer is (E), since nn grows far more rapidly than n!, and 0 ≤
all x. But let’s work this out in case it’s not clear to you.
x2n
1+x2n
< 1 for
From the remarks above, we know
0≤
nn
n!
n!x2n
≤ n,
(1 + x2n )
n
P n!
P n!x2n
for n = 1, 2, 3, . . .. So convergence of
nn implies convergence of
nn (1+x2n ) .
P n!
We will use the ratio test to prove
nn converges.
(n+1)! (n + 1)! nn
(n+1)n+1 =
lim
lim n!
n→∞ n→∞ (n + 1)n+1 n!
nn
(n + 1) nn
n→∞ (n + 1)n+1 1
n
n
= lim
n→∞ n + 1
1
=
e
<1
= lim
t
t
t
t
Before we move on to the next problem, let’s see why limt→∞ t+1
= 1/e. Let y = t+1
. Then, taking
the natural log of both sides and using the power rule for logarithms on the right side of the equation yields,
33
log y = t log
t
t+1 .
It follows that
lim log y = lim t log
t→∞
t→∞
log
t
t+1
t
t+1
1
t
t+1
1
LH
t · (t+1)2
= lim
t→∞
− t12
2
= lim
t→∞
−t
+t
−1
= lim
t→∞ 1 + 1
t
= lim
t→∞ t2
= −1.
And lim log y = −1 implies lim y = 1/e. Note “LH” in the string of equations above is short for L’Hôspital’s
t→∞
t→∞
rule. A list of logarithm properties is located in the glossary.
Question 58.
Suppose A and B are n × n invertible matrices, where n > 1, and I is the n × n identity matrix. If A and
B are similar matrices, which of the following statements must be true?
I. A − 2I and B − 2I are similar matrices.
II. A and B have the same trace.
III. A−1 and B −1 are similar matrices.
(A) I only (B) II only (C) III only (D) I and III only
(E) I, II, and III
Solution. Let’s introduce some notation. Since A and B are similar, let’s say P AP −1 = B. Also let tr(C)
denote the trace of C.
Onto our problem. Property I is true. Suppose P AP −1 = B. Then
P (A − 2I)P −1 = P AP −1 − 2P IP −1
= P AP −1 − 2I
= B − 2I.
Property II is valid: Recall tr(AB) = tr(BA). In our case
tr(B) = tr P (AP −1 )
= tr (AP −1 )P
= tr(AI)
= tr(A).
Property III holds too: If P AP −1 = B, then
34
B −1 = (P AP −1 )−1
= (P −1 )−1 (P A)−1
= P A−1 P −1 .
Question 59.
Suppose f is an analytic function of the complex variable z = x + iy given by
f (z) = (2x + 3y) + ig(x, y),
where g(x, y) is a real-valued function of the real variables x and y. If g(2, 3) = 1, then g(7, 3) =
(A) -14 (B) 9 (C) 0 (D) 11 (E) 18
Solution. Recall the necessary and sufficient condition for a function to be analytic:
The function f (z) = u(x, y) + iv(x, y) is analytic if and only if ux (x, y) = vy (x, y) and uy (x, y) =
−vx (x, y).
∂
∂
Hence, if f is analytic, then gy (x, y) = ∂x
(2x + 3y) = 2 and gx (x, y) = − ∂y
(2x + 3y) = −3. It follows that g(x, y) = 2y + h1 (x) and g(x, y) = −3x + h2 (y). In conjunction, these two expressions imply
g(x, y) = −3x + 2y + C, where C is a constant. We know g(2, 3) = −3(2) + 2(3) + C = C = 1. Thus,
g(7, 3) = −3(7) + 2(3) + 1 = −15 + 1 = −14.
Question 60.
The group of symmetries of the regular pentagram shown above is isomorphic to the
(A) symmetric group S5
(B) alternating group A5
(C) cyclic group of order 5
(D) cyclic group of order 10
(E) dihedral group of order 10
Solution. The Dihedral group of order 10 is, by definition, the group of symmetries of a pentagon. There’s
no reason to believe the symmetries of a pentagram are any different.
35
Question 61.
Which of the following sets has the greatest cardinality?
(A) R
(B) The set of all functions from Z to Z
(C) The set of all functions from R to {0, 1}
(D) The set of all finite subsets of R
(E) The set of all polynomials with coefficients in R
Solution. A list of definitions for cardinal arithmetic is contained in the glossary. Also, recall the SchröderBerstein theorem, which states:
If |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|.
Onto our problem.
(A) Recall |R| = 2ℵ0 .
ℵ 0
= 2ℵ0 ·ℵ0 = 2ℵ0 . Thus, by the Schröder-Berstein
(B) {f | f : Z → Z} = ℵℵ0 0 , and 2ℵ0 ≤ ℵℵ0 0 ≤ 2ℵ0
theorem, we conclude {f | f : Z → Z} = 2ℵ0 . Note |Z| = ℵ0 .
ℵ0
(C) It’s clear {f | f : R → {0, 1}} = 22 .
(D) Let F denote the set of finite subsets of R. Because f : R → F defined by f : x 7→ {x} is one-to-one,
it’s clear 2ℵ0 ≤ |F|. Let Fn denotes the set of subsets of R with n elements. Then
|F| =
≤
∞
X
n=0
∞
X
|Fn |
2ℵ0
n=1
= ℵ0 · 2ℵ0
= 2ℵ0 .
So, by the Schröder-Berstein theorem |F| = 2ℵ0 .
(E) It’s clear the set of polynomials has the same cardinality as F.
ℵ0
Since 22
is different than the others, the correct answer must be (C).
Question 62.
Let K be a nonempty subset of Rn , where n > 1. Which of the following statements must be true?
I. If K is compact, then every continuous real-valued function defined on K is bounded.
II. If every continuous real-valued function defined on K is bounded, then K is compact.
III. If K is compact, then K is connected.
(A) I only (B) II only (C) III only
(D) I and II only (E) I, II, and III
Solution. Property I must be true. The argument goes as follows. The continuous image of a compact
space is compact. Thus, for f continuous and real valued, f (K) ⊆ R must be compact. By the Heine-Borel
theorem, f (K) must be closed and bounded.
36
Property II is also true. We’ll prove it using the other direction of the Heine–Borel theorem. If K where
unbounded, then the real valued function f (x) := kxk would be unbounded, where k · k denotes the Euclidian
metric. If K weren’t closed, then it wouldn’t contain one of its limit points, say k. Then f (x) := 1/kx − kk
would be an unbounded continuous function. Thus, since K is a closed and bounded subset of Rn , it must
be compact.
Property III is a lie. Consider [0, 1] ∪ [2, 3]; clearly that’s a compact set and it’s not connected.
Question 63.
If f is the function defined by
(
2
−2
xe−x −x
f (x) =
0
if x 6= 0
if x = 0,
at how many values of x does the graph of f have a horizontal tangent line?
(A) None (B) One (C) Two
(D) Three (E) Four
Solution. The graph of f is horizontal when f 0 (x) = 0. Obviously, this means we need to find the derivative.
2
−2
For x 6= 0, f (x) = xe−x −x , therefore
2
−x−2
2
−x−2
f 0 (x) = e−x
= e−x
=
+ x(−2x + 2x−3 )e−x
1 − 2x2 + 2x−2
2
−x−2
−2x4 + x2 + 2 −x2 −x−2
e
x2
for x 6= 0. It’s easy to see f 0 (x) = 0 occurs when −2x4 + x2 + 2 = 0 which implies x2 =
Since x2 ≥ 0 we exclude
√
1− 17
4
√
−1± 17
−4
=
√
1± 17
.
4
as a possibility. So we’ve found 2 values of x that do the job so far. Namely,
p
√
1 + 17
x=±
.
2
The only other time, possibly, when f 0 (x) = 0 is at x = 0. For this case, we need to go back to the derivative
definition. We have
f (h) − f (0)
h
−h2 −h−2
he
= lim
h→0
h
f 0 (0) = lim
h→0
2
= lim e−h
−h−2
h→0
= 0.
Note −h2 − h−2 → −∞ as h → 0. Thus, there are a total of three values of x that make f parallel to the
x-axis.
37
Question 64.
For each positive integer n, let fn be the function defined on the interval [0, 1] by fn (x) =
of the following statements are true?
xn
. Which
1 + xn
I. The sequence {fn } converges point-wise on [0, 1] to a limit function f .
II. The sequence {fn } converges uniformly on [0, 1] to a limit function f .
Z 1
Z 1
fn (x) dx =
lim fn (x) dx.
III. lim
n→∞
0
0 n→∞
(A) I only (B) III only (C) I and II only
(D) I and III only (E) I, II, and III
Solution. First, it’s clear that
xn
f (x) = lim
=
n→∞ 1 + xn
(
1/2,
0,
x=1
x 6= 1
As such, we can conclude that {fn } converges point-wise, so I is true. However, {fn } does not converge
uniformly to f by the contrapositive of the uniform convergence theorem, which disproves II. Here’s the
uniform convergence theorem (from Wikipedia):
Suppose {fn } is a sequence of continuous functions that converge point-wise to the function f .
If {fn } converges uniformly to f on an interval S, then f is continuous on S.
It’s clear that III is true. One way to make this conclusion is by considering the fact that the set of discontinuities of f has measure 0. To give a more intuitive argument for III, the area under the graph of fn (x)
R1
R1
goes to zero as n goes to infinity, so 0 fn (x) dx → 0 as n → ∞. Clearly, 0 f (x) dx = 0, since there’s no
area under a point.
Question 65.
Which of the following statements are true about the open interval (0, 1) and the closed interval [0, 1]?
I. There is a continuous function from (0, 1) onto [0, 1].
II. There is a continuous function from [0, 1] onto (0, 1).
III. There is a continuous one-to-one function from (0, 1) onto [0, 1].
(A) none
(B) I only (C) II only (D) I and III only (E) I, II, and III
Solution. Statement I is true. Consider f (x) := sin(2πx); f (1/2) = 0, f (1/4) = 1, and every value
between follows from the intermediate value theorem.
Statement II is false. The image of a compact set under a continuous map is compact. It follows that f ([0, 1])
must be compact when f is continuous. But the Heine-Borel theorem implies f ([0, 1]) must be closed and
(0, 1) is open. Thus f ([0, 1]) 6= (0, 1), if f is continuous.
Statement III is also false. Suppose for the sake of contradiction that g : (0, 1) → [0, 1] is one-to-one and
onto. If g is one-to-one, then it must be monotonic. Since g is onto there exists an x1 in (0, 1) such that
g(x1 ) = 1. But this means g must be increasing for values of x less than x1 and decreasing for values greater
than x1 . This contradicts monotonicity.
38
Question 66.
Let R be a ring with a multiplicative identity. If U is an additive subgroup of R such that ur ∈ U for all
u ∈ U and for all r ∈ R, then U is said to be a right ideal of R. If R has exactly two right ideals, which of
the following must be true?
I. R is commutative.
II. R is a division ring (that is, all elements except the additive identity have multiplicative inverses).
III. R is infinite.
(A) I only
(B) II only (C) III only (D) I and II only (E) I, II, and III
Solution. Statement I is false. Consider the quaternions. Since every nonzero element has an inverse, there
are no proper ideals. Furthermore, 1 is an element of the quaternion. So all the criteria listed above are
satisfied. But ij = k and ji = −k.
Statement II is true. Let x be an arbitrary element in R \ {0}. We will prove x has an inverse. Consider
xR := {xr : r ∈ R}. We want to prove that xR is a right ideal. Let’s first verify that xR is a subgroup.
Closed under addition: Suppose xr and xs are in xR. Then xr + xs = x(r + s) is in xR.
Contains the additive identity: This is clear, since x0 = 0.
Contains additive inverses: It’s clear x(−r) is in xR, and x(−r) + xr = xr + x(−r) = x(r − r) =
x0 = 0.
The subgroup xR is a right ideal, because s in R implies rs is in R, which implies (xr)s = x(rs) is in xR. It
follow that xR = R. This is because x = x1 is in xR and the only right ideals are {0} and R. Hence, there
is x0 in R such that xx0 = 1, so every element must have a right inverse because x 6= 0 was arbitrary. The
element x0 is also a left inverses. This is because (x0 x)(x0 x) = x0 (xx0 )x = x0 x and multiplying both sides by
the right inverse of x0 x proves x0 x = 1. Thus every nonzero element has an inverse and we conclude that R
is a division ring.
Statement III is false. Consider Z3 . It’s a ring, and Z3 only has trivial ideals, since it’s a field.
Yeah, we’re done!
39
GRE mathematics subject test GR9768 solutions are on sale
Solutions to the math subject test GR9768 are on sale for $3 at http://www.rambotutoring.com/booklets.
php. In addition to 45 pages within the main text, there is a 6 page glossary which contains important
definitions and theorems. In total, the document is 53 pages. The formatting is very similar to the solutions
above. If you liked the GR0568 solutions, I think you’ll like my GR9768 solutions too, because I really upped
my game. The original questions to the math subject test GR9768 can be viewed at http://www.math.
ucla.edu/~cmarshak/GRE4.pdf.
A paperback containing both sets of solutions is on sale
A paperback booklet, containing the above GR0568
solutions and the GR9768 solutions, is on sale for
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the author is yours truly (Charles Rambo), and you
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same to you.
40
Glossary
Antiderivatives Useful antiderivatives.
Z
un+1
•
un du =
+ C, n 6= −1
n+1
Z
•
eu du = eu + C
Z
•
du
= log |u| + C
u
Z
•
sin u du = − cos u + C
Z
•
cos u du = sin u + C
Z
•
tan u du = − log | cos u| + C
Z
•
du
= Arctan u + C
1 + u2
Arc length
• Curve described in terms of a function: Suppose y = f (x). Then the arc length from x = a to
x = b is
s
2
Z b
dx
1+
dx.
dt
a
• Curve described in terms of a third parameter: Suppose x = f (t) and y = g(t) describe a curve.
Then the arc length from t = a to t = b is
s
Z b 2 2
dx
dy
+
dt.
dt
dt
a
• Curve described in polar coordinates: Suppose r = f (θ). Then the arc length from θ = α to
θ = β is
s
2
Z β
dr
r2 +
dθ.
dθ
α
41
Basis The set B is a basis of a vector space V over a field F if and only if
• B is nonempty,
• every element in V can be written as a linear combination of elements in B,
• the elements of B are linearly independent.
Binomial distribution Suppose n independent trials are conducted, each of which can either end in success
or failure. Let p be the probability success. Then the probability of exactly k trials ending in success
is
n k
p (1 − p)n−k .
k
Furthermore, in a binomial distribution
• The mean is µ = np
• The variance is σ 2 = np(1 − p)
• The standard deviation is σ =
p
np(1 − p)
Cardinal arithmetic Let A and B be sets.
`
`
• |A| + |B| = |A B| where A B denotes the disjoint union of A and B.
• |A||B| = |A × B|.
• {f |f : A → B} = |B||A| .
• |A| · |B| = sup{|A|, |B|} if |A| or |B| is an infinite cardinal.
Compact Consider the set X under some topology. A collection U of open sets is said to be an “open
cover” if and only if
[
X⊆
U.
U∈ U
The set X is compact if and only if every open cover U has a finite subcover {U1 , U2 , . . . , Un } ⊆ U such
that
X ⊆ U1 ∪ U2 ∪ . . . ∪ Un .
Dense Suppose A and B are subsets of a topological space X for a definition of topological space if you
don’t know this term). We say A is dense in B if and only if cl(A) = B. In other words, every point
of B is a limit point of A or an element of A.
Descartes’s rule of signs Suppose f (x) = an xn + an−1 xn−1 + . . . + a1 x + a0 . Then the number of positive
zeros of f is equal to the number of sign changes of f (x) or is an even number less. Furthermore, the
number of negative zeros of f is equal to the number of sign changes of f (−x) or is an even number
less.
e
e :=
n
∞
X
1
n+1
= lim
.
k! n→∞
n
k=0
Euler’s formula eiθ = cos θ + i sin θ for all θ in C.
42
First derivative test Suppose f : R → R is continuous on the open interval (a, b) and differentiable on
(a, b) \ {c}.
• If f 0 (x) > 0 for x in (a, c) and f 0 (x) < 0 for x in (c, b), then f (c) is a relative maximum.
• If f 0 (x) < 0 for x in (a, c) and f 0 (x) > 0 for x in (c, b), then f (c) is a relative minimum.
Fundamental counting principle Suppose there are n1 ways to complete one activity, and n2 ways of
completing another independent activity. Then there are
n1 · n2
ways to complete both. More generally, if there are ni ways to complete the i-th independent activity,
where i = 1, 2, . . . , m then there are
m
Y
ni
i=1
ways to complete all m activities.
Fundamental theorem of Calculus Suppose f is continuous on [a, b]. Then
Z
b
f (x) dx = F (b) − F (a),
a
where F 0 (x) = f (x).
Fundamental theorem of finitely generated abelian groups Let G be a finitely generated abelian
group. Then it is isomorphic to an expression of the form
Zk × Zpα1 1 × Zpα2 2 × . . . Zpαnm ,
where k, α1 , α2 , . . . , αm are whole number and p1 , p2 , . . . , pm are primes which are not necessarily
distinct. Alternatively, G is isomorphic to an expression of the form
Zk × Zr1 × Zr2 × . . . × Zrn ,
where k, r1 , r2 , . . . , rn are whole numbers and ri divides ri+1 for all i = 1, 2, . . . , n − 1. Note that k and
each ri are uniquely determined by G.
Green’s theorem Let C be a positively oriented, piecewise smooth, simple closed curve in the xy-plane,
and let D be the region bounded by C. Suppose L and M are functions of x and y and have continuous
partial derivatives on an open region containing D, then
I
ZZ
∂L
∂M
−
dA,
L dx + M dy =
∂x
∂y
C
D
where the path of integration along C is counterclockwise. http://en.wikipedia.org/wiki/Green_
theorem
Group The set G together with a binary operation · is a group if and only if
• a and b in G implies a · b in G;
• for all a, b, and c in G, we have (a · b) · c = a · (b · c);
• there is an element e such that e · a = a · e = a for all a in G;
• and for all a in G there is a−1 such that a · a−1 = a−1 · a = e.
43
http://en.wikipedia.org/wiki/Group_(mathematics)
Heine-Borel theorem A set in Rn is closed and bounded if and only if it is compact.
Integration properties Suppose f and g are integrable real valued functions over [a, b]. Let α and β be
in R. Then
Z b
Z b
Z b
g(x) dx
f (x) dx + β
αf (x) + βg(x) dx = α
•
b
•
a
Z
f (x) dx = −
f (x) dx
a
Z
•
a
a
a
Z
b
b
Z
Z
a
b
f (x) dx
f (x) dx +
f (x) dx =
a
c
c
Z
• If f (x) ≤ g(x) for x in [a, b], then
b
Z
f (x) dx ≤
a
b
g(x) dx
a
Intermediate value theorem Suppose f is a real-valued continuous function on the interval [a, b]. For
each y between f (a) and f (b), there is a c in [a, b] such that f (c) = y.
L’Hôspital’s rule Let f and g be functions differentiable on (a, b)\{c}, and g(x) 6= 0 for all x in (a, b)\{c},
where c is in (a, b). Assume limx→c f (x) = limx→c g(x) = 0 or limx→c f (x) = limx→c g(x) = ±∞. Then
f (x)
f 0 (x)
= lim 0
.
x→c g(x)
x→c g (x)
lim
Lagrange’s theorem For any finite group G, the order of every subgroup H of G divides the order of G.
http://en.wikipedia.org/wiki/Lagrange’s_theorem_(group_theory)
Law of cosines Consider 4ABC shown below.
B
c
a
A
b
C
Then
• a2 = b2 + c2 − 2bc cos A
• b2 = a2 + c2 − 2ac cos B
• c2 = a2 + b2 − 2ab cos C
Logarithm properties The GRE assumes log is base e not base 10.
Z
du
•
= log |u| + C
u
• log x = y ⇐⇒ ey = x
• log 1 = 0
44
• log e = 1
• log(xy) = log x + log y
• log(x/y) = log x − log y
• log xy = y log x
Mean value theorem Suppose f is continuous on [a, b] and differentiable on (a, b). Then there is some c
in (a, b) such that
f (b) − f (a)
f 0 (c) =
.
b−a
http://en.wikipedia.org/wiki/Mean_value_theorem
Method of Lagrange multipliers Suppose f (x, y, z) and g(x, y, z) have continuous first order partial
derivatives, and there is a constant k such that g(x, y, z) = k. Then relative extrema of f occur
at the points (x, y, z) that satisfy
fx (x, y, x) = λgx (x, y, z),
fy (x, y, z) = λgy (x, y, z),
and fz (x, y, z) = λgz (x, y, z)
for some λ in R.
Metric A metric on a set X is a function d : X × X → R such that for all x, y, and z in X the following
hold:
• Non-negativity: d(x, y) ≥ 0 and d(x, y) = 0 if and only if x = y
• Symmetry: d(x, y) = d(y, x)
• Triangle inequality: d(x, z) ≤ d(x, y) + d(y, z).
Necessary and sufficient condition for a function to be analytic The function f (z) = u(x, y)+iv(x, y)
is analytic if and only if
∂v
∂u
∂v
∂u
=
and
=− .
∂x
∂y
∂y
∂x
Newton’s binomial theorem
n
(x + y) =
n X
n
k=0
k
xn−k y k .
Pigeonhole principle Suppose we want to put n items into m slots. Then there must be at least one slot
that contains at least dn/me items and there must be a slot that holds no more than bn/mc.
Probability properties Let X be the sample space, and A and B be events in X.
• P (X) = 1
• P (∅) = 0
• 0 ≤ P (A) ≤ 1
• P (X \ A) = 1 − P (A)
• P (B) ≤ P (A) if B ⊆ A
• P (A \ B) = P (A) − P (A ∩ B)
45
• P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
• P (A ∩ B) = P (A) · P (B) if A and B are independent events
Pythagorean identities Suppose θ in R. Then
cos2 θ + sin2 θ = 1,
1 + tan2 θ = sec2 θ,
and 1 + cot2 θ = csc2 θ
Rank nullity theorem Suppose V is a finite dimensional vector space and let T : V → W be a linear
map. Then
nullity(T ) + rank(T ) = dim(V )
∞
X
an+1 . Then
Ratio test Consider the series S :=
an and the limit L := lim n→∞
a
n
n=1
• if L < 1 then the S converges absolutely.
• if L > 1 then S does not converge.
• if L = 1 or L doesn’t exist, then the test is inconclusive.
Ring A set R which is an abelian group under +, and
• (a · b) · c = a · (b · c) for all a, b, and c in R (i.e. · is associative).
• There is an element 1 in R such that a · 1 = a and 1 · a = a (i.e. there is a multiplicative identity).
• a · (b + c) = (a · b) + (a · c) for all a, b, and c in R (i.e. left distributivity holds).
• (b + c) · a = (b · a) + (c · a) for all a, b, and c in R (i.e. right distributivity holds).
http://en.wikipedia.org/wiki/Ring_(mathematics)
Schröder-Berstein theorem If |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|.
Second derivatives test Suppose that the function f : R2 → R has continuous second order partial
derivatives in some E ⊆ R2 . Suppose the point (a, b) in E is a critical point, i.e. fx (a, b) = 0 and
fy (a, b) = 0. Define
fxx (x, y) fxy (x, y)
2
Hf (x, y) := det
= fxx (x, y)fyy (x, y) − [fxy (x, y)] .
fyx (x, y) fyy (x, y)
• If fxx (a, b) > 0 and Hf (a, b) > 0, then f (a, b) is a relative minimum.
• If fxx (a, b) < 0 and Hf (a, b) > 0, then f (a, b) is a relative maximum.
• If Hf (a, b) < 0, then (a, b) is a saddle point.
• If Hf (a, b) = 0, then the test gives no information.
http://en.wikipedia.org/wiki/Second_partial_derivative_test
Sine and cosine values in quadrant I To convert the radian measures in the first row to degrees, simply
multiply 180◦ /π.
θ
cos θ
sin θ
0
1
0
√π/6
3/2
1/2
46
√π/4
√2/2
2/2
π/3
√1/2
3/2
π/2
0
1
Uniform continuity Consider the metric spaces (X, ρ) and (Y, σ). A function f : X → Y is uniformly
continuous on U ⊆ X if and only if for all ε > 0 there is a δ > 0 such that
σ (f (x1 ), f (x2 )) < ε whenever ρ(x1 , x2 ) < δ,
for all x1 and x2 in U .
Uniform convergence theorem Suppose {fn } is a sequence of continuous functions that converge pointwise to the function f . If {fn } converges uniformly to f on an interval S, then f is continuous on S.
See http://en.wikipedia.org/wiki/Uniform_convergence.
Work Let C := {γ(t) : a ≤ t ≤ b}, where γ : R → R3 is differentiable in each coordinate. Then the work
done by a vector field F over C is
Z
Z
F · dγ =
W =
a
C
47
b
F · γ 0 (t) dt.